A first order reaction has a rate constant of 0.981 sec-1 at 26.7 oC. If the activation energy is 61 kJ, calculate the temperature in oC at which the rate constant is 0.425 sec-1. Use 1 decimal place

T2 = 289.8 K

Explanation:

K1 =  0.981 sec-1

T1 = 26.7 oC = 299.7 K (Converting to Kelvin temperature)

Ea = 61kJ = 61,000 J

K2 = 0.425 sec-1

T2 = ?

The equation relating these parameters is;

ln(K2 / K) = - Ea/R (1 / T2 - 1 / T1)

R is the gas constant (8.314 J/mol-K)

ln (0.425 / 0.981) = - 61,000 / 8.314 (1 / T2 - 1 / T1 )

ln (0.4332) = -7337.02 (1 / T2 - 1 / T1 )

- 0.83656 = - 7337.02 (1 / T2 - 1 / T1 )

0.0001140 = ( 1 / T2 - 1 / 299.7 )

0.0001140 = 1 / T2 - 0.00333667

1 / T2 = 0.0001140 + 0.00333667

1 / T2 = 0.00345067

T2 = 289.8 K

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.